Question Number 85384 by sakeefhasan05@gmail.com last updated on 21/Mar/20
Commented by mathmax by abdo last updated on 21/Mar/20
![let A =∫_0 ^1 (dx/(1+x^4 )) ⇒ A = ∫_0 ^1 (dx/((x^2 +1)^2 −2x^2 )) =∫_0 ^1 (dx/((x^2 +1+(√2)x)(x^2 +1−(√2)x))) F(x) =(1/((x^2 +x(√2)+1)(x^2 −x(√2)+1))) =((ax+b)/(x^2 +x(√2)+1)) +((cx+d)/(x^2 −x(√2) +1)) F(−x)=F(x) ⇒((−ax +b)/(x^2 −x(√2) +1)) +((−cx +d)/(x^2 +x(√2) +1)) =F(x) ⇒ c=−a and d=b ⇒ F(x) =((ax+b)/(x^2 −x(√2)+1)) +((−ax +b)/(x^2 +x(√2)+1)) F(0) =1 =2b ⇒b=(1/2) F(1) =(1/2) =((a+b)/(2−(√2))) +((−a+b)/(2+(√2))) =(((2+(√2))a+(2+(√2))b−(2−(√2))a+(2−(√2))b)/2) ⇒1 =2(√2)a +4.(1/2) =2(√2)a +2 ⇒a=−(1/(2(√2))) ⇒ F(x)=((−(1/(2(√2)))x+(1/2))/(x^2 −x(√2) +1)) +(((1/(2(√2)))x +(1/2))/(x^2 +x(√2) +1)) =(1/(2(√2))){ ((x+(√2))/(x^2 +x(√2) +1)) −((x−(√2))/(x^2 −x(√2) +1))} ⇒ ∫_0 ^1 F(x)dx =(1/(4(√2)))∫_0 ^1 ((2x+2(√2))/(x^2 +x(√2) +1))dx−(1/(4(√2)))∫_0 ^1 ((2x−2(√2))/(x^2 −x(√2) +1))dx} =(1/(4(√2)))[ln(x^2 +x(√2) +1)]_0 ^1 +(1/4)∫_0 ^1 (dx/(x^2 +x(√2) +1))−(1/(4(√2)))[ln(x^2 −x(√2) +1)]_0 ^1 +(1/4)∫_0 ^1 (dx/(x^2 +x(√2) +1)) =(1/(4(√2)))ln(2+(√2))−(1/(4(√2)))ln(2−(√2)) +(1/4){ ∫_0 ^1 (dx/(x^2 +x(√2)+1))+∫_0 ^1 (dx/(x^2 −x(√2) +1))} ∫_0 ^1 (dx/(x^2 +x(√2)+1)) =∫_0 ^1 (dx/(x^2 +2x((√2)/2) +(1/2)+(1/2))) =∫_0 ^1 (dx/((x+(1/( (√2))))^2 +(1/2))) =_(x+(1/( (√2))) =(1/( (√2)))u) ∫_1 ^((√2)+1) (du/( (√2)×(1/2)(1+u^2 ))) =(√2) [arctanu]_1 ^((√2) +1) =(√2){arctan((√2)+1)−(π/4)} we do the same manner for the other integral..](https://www.tinkutara.com/question/Q85407.png)
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:\Rightarrow\:{A}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}+\sqrt{\mathrm{2}}{x}\right)\left({x}^{\mathrm{2}} \:+\mathrm{1}−\sqrt{\mathrm{2}}{x}\right)} \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{x}\sqrt{\mathrm{2}}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}+\mathrm{1}\right)}\:=\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}+\mathrm{1}}\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}\:+\mathrm{1}} \\ $$$${F}\left(−{x}\right)={F}\left({x}\right)\:\Rightarrow\frac{−{ax}\:+{b}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\:+\frac{−{cx}\:+{d}}{{x}^{\mathrm{2}} +{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\:={F}\left({x}\right)\:\Rightarrow \\ $$$${c}=−{a}\:{and}\:{d}={b}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{{ax}+{b}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}+\mathrm{1}}\:+\frac{−{ax}\:+{b}}{{x}^{\mathrm{2}} +{x}\sqrt{\mathrm{2}}+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=\mathrm{2}{b}\:\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{{a}+{b}}{\mathrm{2}−\sqrt{\mathrm{2}}}\:+\frac{−{a}+{b}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:=\frac{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){a}+\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){b}−\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){a}+\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){b}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{1}\:=\mathrm{2}\sqrt{\mathrm{2}}{a}\:+\mathrm{4}.\frac{\mathrm{1}}{\mathrm{2}}\:=\mathrm{2}\sqrt{\mathrm{2}}{a}\:+\mathrm{2}\:\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{x}+\frac{\mathrm{1}}{\mathrm{2}}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\:+\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{x}\:+\frac{\mathrm{1}}{\mathrm{2}}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}\:+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\frac{{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\:−\frac{{x}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\right\}\:\Rightarrow \\ $$$$\left.\int_{\mathrm{0}} ^{\mathrm{1}} \:{F}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{x}+\mathrm{2}\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} +{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{x}−\mathrm{2}\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}{dx}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\left[{ln}\left({x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}\:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} +{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\left[{ln}\left({x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}\:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}\:+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left\{\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} +{x}\sqrt{\mathrm{2}}+\mathrm{1}}+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\right\} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=_{{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{u}} \:\:\:\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}+\mathrm{1}} \:\:\frac{{du}}{\:\sqrt{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\sqrt{\mathrm{2}}\:\left[{arctanu}\right]_{\mathrm{1}} ^{\sqrt{\mathrm{2}}\:+\mathrm{1}} \\ $$$$=\sqrt{\mathrm{2}}\left\{{arctan}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)−\frac{\pi}{\mathrm{4}}\right\} \\ $$$${we}\:{do}\:{the}\:{same}\:{manner}\:{for}\:{the}\:{other}\:{integral}.. \\ $$
Answered by MJS last updated on 21/Mar/20
$$\int\frac{{dx}}{{x}^{\mathrm{4}} +\mathrm{1}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\int\frac{{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}−\int\frac{{x}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}\right)= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\int\frac{\mathrm{2}{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}− \\ $$$$\:\:\:\:\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\int\frac{\mathrm{2}{x}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:− \\ $$$$\:\:\:\:\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\left(\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:+\mathrm{2}\left(\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\right)\right)\:+{C} \\ $$